Answer:
the answer is A
Explanation:
According to the law of conservation of energy or the first law thermodynamics energy neither be created nor destroyed, energy is transferred from one form to another form.
Lowery-Bronsted theory is applied here. Acc. to this theory Base accepts protons and Acids donate proton.
Part 1:
Aniline is less basic than ethylamine because the lone pair on nitrogen (which accepts proton) is not localized. It resonates throughout the conjugated system of phenyl ring. Hence due to unavailability of electrons for accepting proton it is less basic compare to ethylamine. In ethyl amine the lone pair of electron is localized and available to abstract proton.
Part 2:
In this case the alkyl groups attached to -NH₂ (in ethylamine) and -O⁻ (in ethoxide are same (i.e. CH₃-CH₂-). Ethoxide is more basic than ethylamine because ethoxide is a conjugate base of ethanol (pKa value of ethanol = 15.9 very weak acid) and the conjugate base of weak acid is always a strong base. Secondly, the oxygen atom more Electronegative than Nitrogen atom can attract more electron cloud from alkyl group as compared to Nitrogen in ethylamine. Hence, oxygen in ethoxide attains greater electron cloud than the nitrogen in ethylamine. Therefore, it is more basic than ethylamine.
Answer:
The element is CARBON
The number 6 refers to the ATOMIC NUMBER
the numbers 12, 13, and 14 refer to the ATOMIC MASS
how many protons and neutrons are in the first isotope?
<u>6</u><u>. </u><u> </u><u> </u><u> </u><u> </u><u>6</u>
how many protons and neutrons are in the second isotope?
<u>6</u><u>. </u><u> </u><u> </u><u> </u><u> </u><u> </u><u>7</u>
<u>how many protons and neutrons are in the </u><u>t</u><u>h</u><u>i</u><u>r</u><u>d</u><u> </u><u>isotope?</u>
<u>6</u><u>. </u><u> </u><u> </u><u> </u><u> </u><u> </u><u> </u><u>8</u>
<u>y</u><u>o</u><u>u</u><u> </u><u>a</u><u>r</u><u>e</u><u> </u><u>w</u><u>e</u><u>l</u><u>c</u><u>o</u><u>m</u><u>e</u><u> </u><u>:</u><u>)</u>
Answer: 600 kJ
-
Explanation:
C₃H₈ (g) + 5 O₂ (g) =============== 3 CO₂ (g) + 4 H₂O (l)
Δ⁰Hf kJ/mol -104 0 -393.5 -285.8
Δ⁰Hcomb C₃H₈ = 3(-393.5) + 4 (-285.80) - (-104) kJ/mol
Δ⁰Hcomb = 2219.70 kJ/mol
n= m /MW MW c₃H₈ = 44.1 g/mol
n= 12 g/44.1 g/mol = 0.27 mol
then for 12 g the heat released will be
0.27 mol x 2219.70 kJ/mol = 600 KJ
Answer:
Volume of the solutions
This is the most important factor for her to control.